Question

$2x + 5y = 25 \\ 3x - 2y = 9$
A) How to solve it by elimination method.

B) by substitution method
C) graphical method ​

Answer 1

I will show you A and B, and give you a hint how to do C, since brainly doesn't provide C option.

A.

The elimination method is how I would solve this system

$

\begin{cases}

2x+5y=25 \\

3x-2y=9

\end{cases}

$

We start by multiplying the first equation by 3 and second by -2 to obtain

$

\begin{cases}

6x+15y=75 \\

-6x+4y=-18

\end{cases}

$

Now add the two equations and thus eliminate the x-terms to obtain

$19y=57\Longrightarrow y=\dfrac{57}{19}=3$

Then we insert the y into either of the original equations and solve for x

$

2x+5\cdot3=25 \\

2x=25-15 \\

2x=10 \\

x=5

$

And we have the solution at $P(5,3)$.

B.

We pick one equation and express x

$

2x+5y=25 \\

2x=25-5y \\

x=\dfrac{25-5y}{2}

$

Then we replace the x in other equation with expression we just calculated and solve for y

$

3\cdot\dfrac{25-5y}{2}-2y=9 \\

\dfrac{75-15y}{2}-2y=9 \\

\dfrac{75-15y-4y}{2}=9 \\

75-19y=18 \\

-19y=-57 \\

y=3

$

Now we put y in either of the original equations to get x

$

2x+5\cdot3=25 \\

2x=25-15 \\

2x=10 \\

x=5 \\

$

And again we get the solution $P(5,3)$.

C.

To find solution graphically, you need to put both lines in general form $f(x)=mx+n$ where $f(x)=y$

$

\begin{cases}

2x+5y=25\Longrightarrow y=\frac{25-2x}{5}=-\frac{2}{5}x+5 \\

3x-2y=9\Longrightarrow y=-\frac{9-3x}{2}=-\frac{3}{2}x-\frac{9}{2}

\end{cases}

$

Then plot the lines on the graph. The solution will be point $P(5,3)$ where lines intersect, because we can represent lines as sets of points ordered pairs. Consider our two lines represented as two functions g and f where

$

f(x)=-\frac{2}{5}x+5 \\

g(x)=-\frac{3}{2}x-\frac{9}{2}

$

Graphs (sets) of these two functions are defined as (assuming $f,g:D_f,D_g\to T_f,T_g=\mathbb{R}\to\mathbb{R}$ since not specified):

$

\displaystyle

G_f=\{(x,f(x)):x\in D_f\wedge f(x)\in T_f\} \\

G_g=\{(x,g(x)):x\in D_g\wedge g(x)\in T_g\}

$

Where D is definition zone (set of all x defined for function f or g) and T transformation zone (set of all f(x) or g(x)). Note that in our case $D_f,T_f=D_g,T_g$ because $D_f,T_f,D_g,T_g=\mathbb{R}$

The solution is is intersection of these two sets

$G_f\cap G_g=(5,3)$

Hope this helps.